Conic Sections Question 521
Question: If $ \theta $ and $ \varphi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , then $ \theta -\varphi $ is equal to
Options:
A) $ \pm \frac{\pi }{2} $
B) $ \pm \pi $
C) 0
D) None of thesew
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y=m_1x $ and $ y=m_2x $ be a pair of conjugate diameters of an ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ and let $ P(a\cos \theta ,b\sin \theta ) $ and $ Q(a\cos \varphi ,b\sin \varphi ) $ be ends of these two diameters. Then $ m_1m_2=-\frac{b^{2}}{a^{2}} $
$ \Rightarrow \frac{b\sin \theta -0}{a\cos \theta -0}\times \frac{b\sin \varphi -0}{a\cos \varphi -0}=-\frac{b^{2}}{a^{2}} $
therefore $ \sin \theta \sin \varphi =-\cos \theta \cos \varphi $
therefore $ \cos (\theta -\varphi )=0\Rightarrow \theta -\varphi =\pm \frac{\pi }{2} $ .
BETA
